Q. 73.6( 12 Votes )

In Fig. 6.14, DE || QR and AP and BP are bisectors of EAB and RBA, respectively. Find APB.


Answer :

It is given to us –


DE || QR and n is the transversal intersecting DE and QR at points A and B respectively.


We have to find APB.


AP and BP are the bisectors of EAB and ABR respectively.


EAP = PAB


EAB = 2 × PAB - - - - (i)


Also, RBP = PBA


RBA = 2 × PBA - - - - (ii)


Since, DE || QR,


RBA = EAn (Corresponding angles)


RBA = 180° - EAB (By linear pair axiom, EAn + EAB = 180°)


Using equations (i) and (ii) in the above equation,


2 × PBA = 180° - (2 × PAB)


PBA = 90° - PAB (Dividing both sides by 2)


PBA + PAB = 90° - - - - (iii)


Now, in ΔPAB,


PBA + PAB + APB = 180° (Sum of three angles of a triangle is 180°)


90° + APB = 180° [From equation (iii)]


APB = 90°


Thus, the value of APB is 90°.


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